In a semicircle, draw a maximum right triangle with the center point o as the right vertex. It is known that the area of the triangle is 25 square meters. How about the area of the circle?

Draw a maximum right triangle with the center point o as the right vertex, then the right triangle is an equilateral right triangle, and the right side is the radius of the circle. Given that the area of the triangle is 25 square meters, then the side length of the right triangle is 5 √ 2 meters
The garden area is: 3.14 × (5 √ 2) &# 178; = 3.14 × 50 = 157 square meters

If the line L1: X-Y + 1 = 0 and L2: 2x-2y-1 = 0 are two tangents of a circle, then the area of the circle

These two straight lines are parallel, so the distance between the two straight lines is the diameter of the circle, that is, the root sign 2 whose diameter D is three fourths, so the area is the square of π which is nine fourths of 32

The line 6x-2y-1 = 0, 3x-y + 4 = 0 are two parallel tangents of a circle, and the area of the circle is

Solution. ∵ two equal lines whose two straight lines are circles
The distance between two straight lines must be the diameter D of the circle
∴d=|-1/2-4|/√(9+1)=9√(10)/20
The area of a circle s = π (D / 2) & sup2; = 81 π / 160

If the line l1:3x-y + 4 = 0 and l2:6x-2y-1 = 0 are two tangents of a circle, then the area of the circle?
Such as the title

The distance between two straight lines is the diameter
Take any point in L1, such as (0,4)
His distance to L2 = | 0-8-1 | / √ (6 & sup2; + 2 & sup2;) = 9 / √ 40
So the diameter is 9 / √ 40
So area = 81 π / 160

Given that the lines 3x-y + 4 = 0 and 6x-2y-1 = 0 are two tangents of a circle, then the area of the circle is equal to?

Let (a, b) be the center of the circle
|3a-b+4|/√(3^2+1)=|6a-2b-1|√(6^2+2^2)
|3a-b+4|/√10=|6a-2b-1|/√40
2|(3a-b)+4|=|2(3a-b)-1|
|2(3a-b)+8|=|2(3a-b)-1|
2(3a-b)+8=-2(3a-b)+1
4(3a-b)=-7
(3a-b)=-7/4
r=|-7/4+4|/√10=21/4√10
Area = π (21 / 4 √ 10) ^ 2 = 441 / 160 π
I don't know if it's wrong, but that's the way of thinking

If a tangent is drawn from a point on the straight line y = x + 1 to the circle (x-3) 2 + y2 = 1, the minimum length of the tangent is ()
A. 1B. 22C. 7D. 3

The minimum value of tangent length is obtained when the distance between the point on the straight line y = x + 1 and the center of the circle is the smallest, the distance between the center of the circle (3,0) and the straight line is d = | 3 − 0 + 1 − 2 = 22, the radius of the circle is 1, so the minimum value of tangent length is D2 − R2 = 8 − 1 = 7, so select C

If a tangent is drawn from a point on the line y = x + 1 to a circle (x-3) 2 + (y + 2) 2 = 1, the minimum length of the tangent is ()
A. 17B. 32C. 19D. 25

To minimize the tangent length, the distance between the point on the line y = x + 1 and the center of the circle should be the shortest. This minimum value is the distance d from the center of the circle (3, - 2) to the line y = x + 1, d = | 3 + 2 + 1 | 2 = 32, so the minimum tangent length is D2 − R2 = 18 − 1 = 17, so select a

If a tangent is drawn from a point on the straight line y = x + 1 to the circle (x-3) 2 + y2 = 1, the minimum length of the tangent is ()
A. 1B. 22C. 7D. 3

If a tangent is drawn from a point P (2,3) outside the circle (x-1) 2 + (Y-1) 2 = 1, the length of the tangent is______ ．

If the center of the circle is point C and the center of the circle is (1,1), then | PC | 2 = (2-1) 2 + (3-1) 2 = 5. By using the distance formula between two points, it is obtained that | PC | 2 = (2-1) 2 + (3-1) 2 = 5. According to Pythagorean theorem, the tangent length = | PC | 2 − 1 = 5 & nbsp; − 1 = 2

Through the point P (- 2, - 3), make two tangents of the circle C: (x-4) ^ 2 + (Y-2) ^ 2 = 9, and the tangent bands are a and B respectively;
(1) The equation of the circle passing through the center C, the tangent points a and B; (2) the equation of the straight line AB; (3) the length of the line ab
To process ~! Urgent, online, etc. @ ~ 1 ~~@
How no answer ~!

Let (- 2, - 3) be p, then the center of the circle d must be on the vertical bisector of AB, that is, PC. because Da = DC = dB, and PAC is a right triangle, PC is a hypotenuse, so D is the midpoint of PC, that is, (1, - 1 / 2), and the radius is √ 61 / 2, so the equation is (x-1) & sup2; + (y + 1 / 2) & sup2; = 61 / 4

In a semicircle, draw a maximum right triangle with the center point o as the right vertex. It is known that the area of the triangle is 25 square meters. How about the area of the circle?